Be a real number that satisfying: $\displaystyle a > 1$. Consider the sequence defined by:
$\displaystyle X_1 = 1$; $\displaystyle X_{n+1} = [\frac{X_n}{2} + \frac{a}{2x_n}]$
1) Prove that $\displaystyle X_2 > X_1$
As $\displaystyle X_{n+1} = [\frac{X_n}{2} + \frac{a}{2X_n}]$
Then $\displaystyle X_{2} = [\frac{X_1}{2} + \frac{a}{2X_1}]$
$\displaystyle X_1 = 1$ so
$\displaystyle X_{2} = [\frac{1}{2} + \frac{a}{2\times 1}]$
$\displaystyle X_{2} = [\frac{1+a}{2}]$
If $\displaystyle a=1$ then $\displaystyle X_{2} = \frac{1+1}{2}=1$
as $\displaystyle a>1 $ this implies what?
Thank you.
2) Show that if $\displaystyle n \ge 2$, $\displaystyle X_n > X_{n-1}$ then $\displaystyle X_{n+1} > X_n$
It is?
$\displaystyle X_1 = 1 ==> n = 1$
$\displaystyle X_2 = 2 ==> n = 2$
$\displaystyle X_n = n$
$\displaystyle X_{n-1} = n-1$
$\displaystyle X_{n+1} = n + 1$
?
Thanks
3) Prove that $\displaystyle X_1 < a$
I can use my solution in problem 1 ?
$\displaystyle X_2 = \frac{1+a}{2}$
$\displaystyle a = 2X_2-1$
$\displaystyle X_2>X_1$ --> $\displaystyle X_1=1 ==> X_2>1$
if $\displaystyle X_2=1 ==> a=1$ then $\displaystyle X_2>1 ==> a>1$
$\displaystyle a>X_1$
Correct ?